Chapter 25: DC Motor Drives — Power Electronics

Where this sits

Part 6 · Electric Drives
Chapter 25 of 30 — the first of Part 6

You should already know

Phase control and \(V_{dc}=1.35V_L\cos\alpha\) from Chapter 9, dual converters from Chapter 10, and the chopper classes from Chapter 11.

By the end you can

Draw a DC drive's speed–torque family, size a converter for a given machine, explain why the current loop sits inside the speed loop, and choose a braking method.

Time

≈ 55 min reading · ≈ 50 min problems

i What you'll learn

Why the DC machine was the only practical variable-speed drive for a century — torque and flux are separately controllable by construction.
The three equations that describe everything: \( E = k\Phi\omega \), \( T = k\Phi I_a \), \( V = E + I_aR_a \).
Armature-voltage control below base speed and field weakening above it — the constant-torque and constant-power regions.
Why a phase-controlled drive's speed regulation collapses at light load, and what a freewheeling path or continuous conduction fixes.
Why the current loop goes inside the speed loop, and what that arrangement actually protects.
Three ways to brake — regenerative, dynamic and plugging — and when each is right.
Why the DC drive lost, and where it still has not.

Section 25-1

Why the DC Machine Came First

From the 1890s to the 1980s, if you needed a motor whose speed you could control, you used a DC machine. Not because it was cheap — it was not — nor because it was reliable, with its brushes and commutator. You used it because of one structural property that no AC machine offered.

Its torque and its flux are produced by two separate windings, and you can control them independently.

◇Intuition · two knobs that do not interact

Imagine a machine with two dials. Turn the first and the torque changes; turn the second and the speed changes; and neither disturbs the other. That is a control engineer's ideal, and a separately-excited DC machine very nearly is one.

The commutator is what makes it so. Its job is to reverse each armature coil's current at exactly the moment that coil passes from one field pole to the next, so the armature's magnetic field always sits at 90° to the field winding's, whatever the rotor is doing. Torque is greatest when two fields are perpendicular, and the commutator holds them there mechanically.

Three consequences follow, and together they explain a century of practice:

Torque is proportional to armature current — \( T = k\Phi I_a \) — with no angle to worry about and no delay. Control the current and you have controlled the torque.
Speed is proportional to armature voltage, because the back-EMF must balance it: \( E = k\Phi\omega \).
The two are decoupled, so a simple cascade of two loops controls the machine completely.

An induction motor offers none of this. Its rotor current is induced rather than supplied, its torque depends on slip in a strongly non-linear way, and the flux and torque-producing components of its stator current are hopelessly entangled. Making it behave like a DC machine took until the 1970s in theory and the 1990s in affordable silicon — and Chapter 26 is that story.

What the DC machine charged for the privilege: brushes that wear, a commutator that needs turning, carbon dust in the windings, a speed limit set by commutation, and an enclosure that cannot be sealed. Every one of those is a maintenance cost, and it is why the machine lost as soon as the alternative became practical.

🔑

The three equations of a separately-excited DC machine

\[ E_a = k\Phi\,\omega, \qquad T = k\Phi\,I_a, \qquad V_a = E_a + I_aR_a \]

Eliminating \( E_a \) gives the speed directly: \( \omega = \dfrac{V_a - I_aR_a}{k\Phi} \). Everything in this chapter follows from these three lines.

Section recap. The commutator holds the armature and field fluxes perpendicular, so torque follows armature current and speed follows armature voltage, independently. That decoupling made the DC machine the only practical variable-speed drive for a century — and its brushes are why it lost.

Section 25-2

Two Regions of Operation

The speed equation contains two things a drive can change: the armature voltage \( V_a \) and the field flux \( \Phi \). They give two distinct operating regions with quite different characters.

Write the speed–torque relation Substituting \( I_a = T/(k\Phi) \) into the speed equation:

Working

\[ \omega = \frac{V_a}{k\Phi} - \frac{R_a}{(k\Phi)^2}\,T \]

A straight line: a no-load speed set by \( V_a/(k\Phi) \), and a droop set by \( R_a/(k\Phi)^2 \).
Region 1 — below base speed, vary \(V_a\) Hold \( \Phi \) at its rated value and raise \( V_a \) from zero. The whole line shifts upward without changing slope. Since \( T = k\Phi I_a \) and both \( \Phi \) and the maximum \( I_a \) are fixed, the maximum available torque is constant.
Region 2 — above base speed, weaken \(\Phi\) At \( V_a = V_{rated} \) there is nothing left to raise. To go faster, reduce \( \Phi \): the no-load speed \( V_a/(k\Phi) \) rises. But \( T_{max} = k\Phi I_{rated} \) falls in proportion, so

Result

\[ P = T\omega \approx \bigl(k\Phi I_{rated}\bigr)\left(\frac{V_{rated}}{k\Phi}\right) = V_{rated}I_{rated} = \text{constant} \]

Constant power. The flux cancels out entirely.
Note where base speed is

Definition

\[ \omega_{base} = \frac{V_{rated} - I_{rated}R_a}{k\Phi_{rated}} \]

The speed at rated voltage, rated flux and rated current — the boundary between the two regions.

Interactive · armature voltage and field weakening

A 40 kW machine: 440 V, 100 A, \( R_a = 0.4\,\Omega \), 1500 rpm at rated conditions. Raise the armature voltage to reach base speed, then weaken the field to go beyond it — and watch the power stay put while the torque falls.

Armature voltage V_a 440 V

Field flux Φ/Φ_rated 1.00

No-load speed1649 rpm

Speed at 100 A1500 rpm

Torque at 100 A255 N·m

Shaft power40.0 kW

—

▲Common mistake

"Weaken the field to get more speed — so weaken it a lot and get a lot of speed."

Wrong Treating field weakening as a free speed multiplier.

Right Torque falls in exact proportion, and the machine becomes progressively less stable and harder to commutate.

Three separate limits bite, in this order:

Torque. \( T_{max} = k\Phi I_{rated} \) falls linearly with flux. At 40% flux the machine produces 40% of its torque, so the load must be one that needs less torque at speed — a winder, a spindle, a traction drive at cruise.
Commutation. The commutator must reverse each coil's current within the time a brush spans one segment, and that time shrinks with speed. Meanwhile the armature reaction distorts the field, moving the neutral plane. Push both together and you get flashover — an arc across the commutator that can destroy the machine in seconds. This is usually the binding limit, and it is why 3:1 or 4:1 field weakening is typical and 6:1 is exotic.
Speed regulation. The droop term is \( R_aT/(k\Phi)^2 \) — it goes as the inverse square of flux. Halve the flux and the same torque produces four times the speed drop. A drive that held 1% regulation at base speed holds 4% at half flux.

And a fourth limit that catches people out. Reducing the field current does not reduce the flux instantly: the field winding has a time constant of 0.5 to 5 seconds, because it is a many-turn winding on a large iron circuit. A drive commanded to accelerate rapidly into field weakening will not follow, and the current loop will hit its limit while the flux is still collapsing.

This asymmetry is why field weakening is used for steady high-speed running and almost never for fast transients. Good drives anticipate it, ramping the field ahead of the speed demand.

Section recap. Below base speed, vary the armature voltage: constant available torque. Above it, weaken the field: constant power, with torque falling as the flux does. Field weakening is limited by commutation before it is limited by torque, and the field's own time constant makes it slow.

Section 25-3

Feeding the Armature

The machine wants an adjustable DC voltage, and Parts 2 and 3 built two ways to make one. Both are used, and the choice follows from the supply.

Phase-controlled and chopper-fed DC drives
Property	Phase-controlled (Ch. 8–10)	Chopper-fed (Ch. 11)
Supply	AC mains	DC — battery, traction line, rectified link
Output voltage	\(1.35V_L\cos\alpha\) (3-φ)	\(D\,V_s\)
Ripple frequency	\(6f_s\) — 300 Hz	Switching frequency — kHz
Current smoothing	Often needs a series reactor	Usually none needed
Four quadrants	Dual converter (two bridges)	Class E H-bridge
Regeneration	Natural — invert with \(\alpha>90°\)	Needs a receptive supply or a brake resistor
Input power factor	Poor, and falls with speed	Set by the front end, not the chopper
Response	Limited to \(1/6f_s\) — about 3 ms	Sub-millisecond
Typical use	Industrial drives on mains	Traction, battery vehicles, servos

◈Why a phase-controlled drive loses regulation at light load

A phase-controlled drive that holds speed beautifully under load can behave badly when the load is removed. The cause is discontinuous conduction, and it is the same mechanism as Chapter 11's — arriving here with a visible mechanical consequence.

What happens. Between firing pulses the converter's output voltage can fall below the machine's back-EMF. With enough current flowing, the armature inductance carries it through. At light load there is not enough, so the current reaches zero and the thyristors turn off. The armature is then disconnected from the converter entirely, and its terminals sit at the back-EMF.

The consequences, all unwelcome:

The average voltage rises above \( 1.35V_L\cos\alpha \), because the terminals spend part of each period at the EMF rather than at the converter's output. So the speed rises — sometimes sharply.
The relationship becomes load-dependent. In continuous conduction, speed depends only on \( \alpha \). In discontinuous conduction it depends on \( \alpha \) and the load, which means the speed loop's gain changes underneath it.
The no-load speed can be far above the intended value — in the extreme, approaching the peak of the supply rather than its controlled average.

Three standard remedies:

A series smoothing reactor. More inductance means the current is carried further between pulses, so continuous conduction extends to lighter loads. It is the traditional answer, and the reactor is a substantial component.
A three-phase converter rather than single-phase. Six pulses per cycle instead of two means the gaps are a third as long, so conduction stays continuous at much lighter loads. This alone is often enough.
A minimum load, or a controller that detects the mode and switches gains.

Why chopper drives do not suffer from this. At a switching frequency of a few kilohertz, the interval between pulses is a fraction of a millisecond — far shorter than the armature's electrical time constant — so the current has no opportunity to decay to zero. This is one of the clearest practical advantages of the chopper drive, and it explains why servo applications essentially never use phase control.

Section recap. Phase control suits a mains supply and regenerates naturally, but its 300 Hz ripple limits response and it loses regulation in discontinuous conduction. A chopper suits a DC supply, needs no smoothing reactor and responds in microseconds, but needs a receptive supply or a brake resistor to regenerate.

Section 25-4

Two Loops, and Why the Current One Is Inside

Every DC drive built since about 1960 has the same control structure: a speed loop whose output is a current demand, and an inner current loop that delivers it. The arrangement is so standard that it is worth asking why it is not done the other way round.

◇Intuition · a driver and an accelerator pedal

A driver holding a speed does not think about fuel injection. They press the pedal, and something faster and more local turns that into the right amount of fuel — while also making sure the engine is never asked for more than it can safely deliver.

The speed loop is the driver; the current loop is the engine management. And the arrangement earns its place for four separate reasons.

Current is torque. \( T = k\Phi I_a \), so commanding a current is commanding a torque directly. The speed loop's output is therefore a physically meaningful quantity, not an abstraction.
Protection becomes trivial. Clamp the speed loop's output at \( I_{max} \) and the machine can never draw more, whatever the speed error. During a hard acceleration or a stall, the drive simply delivers full torque and no more. This single property is the main reason for the structure — without it, protection would have to be a separate trip that interrupts operation rather than limiting it.
The two loops have very different speeds. The electrical time constant \( L_a/R_a \) is a few milliseconds; the mechanical time constant \( J\omega/T \) is often hundreds. A fast inner loop and a slow outer loop are almost independent, so each can be tuned on its own.
The inner loop hides the plant. Once the current loop is closed and fast, the speed loop sees something that produces the torque it asks for. Supply variation, back-EMF and armature resistance are all rejected by the inner loop before the outer one notices them.

What happens if you invert them. A current loop outside a speed loop is meaningless — you would be asking the machine to hold a current by adjusting its speed, which is not something a load permits. And a single speed loop with no current loop has no way to limit torque except by tripping.

🔑

The cascade, and its two time constants

\[ \tau_{elec} = \frac{L_a}{R_a} \quad (\text{2–20 ms}), \qquad \tau_{mech} = \frac{J R_a}{(k\Phi)^2} \quad (\text{50–1000 ms}) \]

Tune the current loop for a bandwidth of roughly \( 1/\tau_{elec} \), then the speed loop a factor of five to ten slower. The separation is what makes independent tuning valid.

▣In practice · what the current limit actually does

Follow a drive through a hard start, and the cascade's behaviour becomes concrete.

At \( t = 0 \) the speed reference steps from 0 to 1500 rpm. The speed error is enormous, so the speed controller's output saturates immediately at \( I_{max} \) — typically 150–200% of rated current.
The current loop delivers that current within a few milliseconds, and holds it. The machine produces constant maximum torque.
Constant torque means constant acceleration, so the speed ramps linearly. The drive is now effectively in open-loop torque control, and the speed loop is doing nothing but holding its output at the limit.
As the speed approaches the reference the error falls, the speed controller comes out of saturation, and normal closed-loop regulation resumes.

Three things this structure gives for free:

The acceleration is controlled and repeatable — set by \( I_{max}k\Phi/J \), not by how aggressively the speed loop happens to be tuned.
A stalled motor draws \( I_{max} \), not \( V/R_a \). For this machine that is 150 A rather than \( 440/0.4 = 1100 \) A. The difference is between a drive that limits and one that explodes.
Adjusting the acceleration rate is one parameter. Lower \( I_{max} \) and the machine accelerates more gently, with no retuning.

And the classic failure if anti-windup is omitted. While the speed controller is saturated its integrator continues accumulating error. By the time the speed arrives, the integral term is enormous, so the controller keeps demanding maximum current well past the target — a large overshoot, then a long recovery. Every practical implementation clamps or freezes the integrator while the output is saturated, and it is the first thing to check when a drive overshoots on every start.

Section recap. The speed loop commands a current, and the inner current loop delivers it. Clamping the speed loop's output gives torque limiting by construction, the two time constants differ by an order of magnitude so the loops can be tuned independently, and the inner loop hides the plant from the outer one. Anti-windup is mandatory.

Section 25-5

Three Ways to Stop

A motor accelerating draws power. A motor decelerating produces it, and that energy has to go somewhere. The three answers differ in where it goes, and they are not interchangeable.

Braking methods compared
Method	How it works	Energy goes to	Needs	Best for
Regenerative	Machine becomes a generator; converter inverts	Back to the supply	A receptive supply, or a dual converter / four-quadrant chopper	Frequent braking; hoists; traction
Dynamic (rheostatic)	Armature switched onto a resistor	The resistor, as heat	A resistor and a contactor or chopper	Occasional stops; non-receptive supplies
Plugging	Armature voltage reversed while still turning	The machine and the supply, both	A current-limiting resistor; robust machine	Emergency stops only

▲Common mistake

"Plugging stops the motor fastest, so use it whenever a quick stop is wanted."

Wrong Reversing the armature voltage on a running machine without a current-limiting resistor.

Right The back-EMF and the reversed supply add, so the driving voltage is nearly \( 2V \) across only \( R_a \).

Put numbers on it. The machine of Section 25-2 running at rated speed has \( E = 400 \) V. Reverse the armature supply to −440 V:

\[ I = \frac{V + E}{R_a} = \frac{440 + 400}{0.4} = 2100\ \text{A} \]

Twenty-one times rated current. The commutator would flash over, the armature winding would be thrown out of its slots by the magnetic forces, and the converter would fail — all within a few tens of milliseconds.

So plugging always requires a series resistor, sized to limit the current to about twice rated:

\[ R_{total} \ge \frac{V + E}{2I_{rated}} = \frac{840}{200} = 4.2\ \Omega \;\Longrightarrow\; R_{plug} = 3.8\ \Omega \]

And the energy is worse than it looks. During plugging the supply is also pushing current through the machine, so the resistor dissipates the stored kinetic energy plus whatever the supply delivers — roughly twice the energy that dynamic braking would.

One further trap: the machine does not stop at zero speed. If nothing disconnects it, the reversed voltage accelerates it backwards. A zero-speed detector and a contactor are mandatory.

Which is why plugging is reserved for emergency stops — where a very fast deceleration justifies the stress, the wasted energy and the extra hardware. For routine stopping it is the wrong answer in every respect.

✦Think about it

A drive uses a three-phase dual converter and can therefore regenerate. During a factory shutdown the drive trips on overvoltage every time it decelerates. What is happening?

Regeneration needs somewhere for the energy to go

A dual converter can invert — set \( \alpha > 90^\circ \) and the DC side voltage reverses, so power flows from the machine back through the bridge towards the supply. That much is correct.

But inverting requires the supply to accept the power. During a shutdown, most of the factory's load is off. With little else drawing power, the regenerated energy has nowhere to go: it charges the supply-side capacitance and raises the local voltage until the drive's overvoltage protection trips.

Three situations where regeneration fails for this reason:

A lightly-loaded site. The regenerating drive is a source, and if it is the largest thing connected, the voltage rises.
A generator supply. An engine-driven alternator cannot absorb reverse power at all — it would have to motor the engine.
A diode rectifier front end. Common on chopper drives: diodes cannot pass current backwards, so the DC link simply charges up. This is the most common case by far.

The standard remedies:

A braking chopper — a resistor and a switch across the DC link, turned on when the voltage exceeds a threshold (Chapter 11). Simple, reliable, and wastes the energy as heat.
An active front end — replace the diode rectifier with a PWM converter that can return power to the mains. Efficient and gives unity power factor, but costs roughly as much again as the drive.
A shared DC link — connect several drives' DC links together so a decelerating drive feeds an accelerating one. Very effective on machines with complementary motions, and cheap.
Slow the deceleration ramp so the regenerated power stays below what the site can absorb. Free, and often sufficient.

The lesson worth generalising: "capable of regeneration" is a property of the converter. Whether regeneration actually works is a property of the system, and it must be checked separately.

Section recap. Regenerative braking returns energy to the supply but needs the supply to accept it; dynamic braking burns it in a resistor and always works; plugging is fastest and stresses everything, so it belongs to emergency stops only. Check the supply's receptiveness before relying on regeneration.

Section 25-6

Worked Examples

1 Sizing a phase-controlled drive

Problem. A 440 V, 100 A, 1500 rpm separately-excited machine with \( R_a = 0.4\ \Omega \) is to be driven from a 415 V, 50 Hz three-phase supply. Find the firing angle for rated operation, and the angle needed for 750 rpm at rated torque.

Machine constant

\[ E_{rated} = 440 - 100(0.4) = 400\ \text{V}, \qquad \omega = \frac{2\pi(1500)}{60} = 157.1\ \text{rad/s} \]

\[ k\Phi = \frac{400}{157.1} = 2.546\ \text{V·s/rad}, \qquad T_{rated} = 2.546(100) = 254.6\ \text{N·m} \]

Firing angle at rated speed

\[ V_{dc} = 1.35\,V_L\cos\alpha = 440 \;\Longrightarrow\; \cos\alpha = \frac{440}{1.35(415)} = 0.785 \;\Longrightarrow\; \alpha = 38.2^\circ \]

At 750 rpm, rated torque (so rated current)

\[ E = 2.546\left(\frac{2\pi(750)}{60}\right) = 200\ \text{V}, \qquad V_a = 200 + 40 = 240\ \text{V} \]

\[ \cos\alpha = \frac{240}{560} = 0.428 \;\Longrightarrow\; \alpha = 64.6^\circ \]

Check the headroom. The bridge can produce up to \( 1.35(415) = 560 \) V, and rated operation needs 440 V — so \( \alpha \) never falls below 38°. That margin is deliberate and necessary: it covers supply undervoltage, commutation overlap (Chapter 10), and the extra voltage needed to force current changes during transients.

And note the power factor consequence. The displacement factor is roughly \( \cos\alpha \), so at rated speed it is 0.79 and at half speed 0.43. A phase-controlled DC drive has a power factor that falls with speed, which is one of the strongest arguments against it for a drive that runs slowly for long periods.

2 Field weakening to twice base speed

Problem. The same machine must run at 3000 rpm. Find the flux required, the torque available at rated current, the shaft power, and comment on the speed regulation.

Flux required

\[ \omega = \frac{2\pi(3000)}{60} = 314.2\ \text{rad/s}, \qquad E = V_a - I_aR_a = 440 - 40 = 400\ \text{V} \]

\[ k\Phi = \frac{400}{314.2} = 1.273 \;\Longrightarrow\; \frac{\Phi}{\Phi_{rated}} = \frac{1.273}{2.546} = \textbf{0.50} \]

Torque and power

\[ T = 1.273(100) = 127.3\ \text{N·m} \quad (\text{half of rated}) \]

\[ P = 127.3(314.2) = 40.0\ \text{kW} \quad (\text{unchanged}) \]

Constant power confirmed — twice the speed at half the torque.

Now the regulation, which is the part that surprises people. The speed drop from no load to rated torque is \( \Delta\omega = TR_a/(k\Phi)^2 \):

Working

\[ \text{At full flux:} \quad \Delta\omega = \frac{254.6(0.4)}{2.546^2} = 15.7\ \text{rad/s} = 150\ \text{rpm} \quad (10\% \text{ of } 1500) \]

\[ \text{At half flux:} \quad \Delta\omega = \frac{127.3(0.4)}{1.273^2} = 31.4\ \text{rad/s} = 300\ \text{rpm} \quad (10\% \text{ of } 3000) \]

The same percentage, but twice the absolute droop — because the torque halved while \( 1/(k\Phi)^2 \) quadrupled. Open-loop, a field-weakened machine is noticeably less stiff in absolute terms, which matters for a load whose torque fluctuates.

The closed-loop answer is that the speed loop corrects all of it — but it must correct twice as much, so the loop works harder and the transient dip after a load step is larger.

3 Sizing a dynamic braking resistor

Problem. The 40 kW drive must stop a total inertia of 8 kg·m² from 1500 rpm in 4 seconds using dynamic braking. Find the resistor value, its peak and average power, and the energy it absorbs.

Kinetic energy to dissipate

\[ W = \tfrac12 J\omega^2 = \tfrac12(8)(157.1)^2 = 98.7\ \text{kJ} \]

Average braking power

\[ P_{avg} = \frac{98.7\times10^3}{4} = 24.7\ \text{kW} \]

Resistor value, limiting the initial current to 150 A

\[ R_b = \frac{E_{initial}}{I_{max}} - R_a = \frac{400}{150} - 0.4 = 2.27\ \Omega \]

Peak power in the resistor

\[ P_{peak} = I^2R_b = (150)^2(2.27) = 51\ \text{kW} \]

Read the two power figures together. The peak is 51 kW but the average over the stop is 24.7 kW, and if the machine stops only occasionally the continuous rating needed is far lower still.

So the resistor is specified by three separate numbers:

Resistance 2.27 Ω, set by the current limit.
Energy 98.7 kJ per stop, which sets its thermal mass — it must absorb this without exceeding its temperature rating.
Continuous power, set by the duty cycle. At one stop every ten minutes, \( 98.7\ \text{kJ}/600\ \text{s} = 165 \) W. At one stop a minute, 1.6 kW.

The mistake to avoid is specifying the resistor for its peak power. A 51 kW continuous resistor would be enormous and largely idle; the correct part is a high-energy, moderate-continuous-power braking resistor, and the duty cycle must be stated when ordering it.

One further check. As the machine slows, \( E \) falls, so the braking current and torque fall with it. Deceleration is therefore not constant — it is exponential, and the last few percent of speed takes a disproportionate time. A drive needing a firm stop adds a mechanical brake for the final part.

4 Diagnosing a runaway at light load

Problem. A single-phase phase-controlled drive holds 1000 rpm accurately under load. With the load removed the speed rises to 1400 rpm and becomes erratic, though the firing angle has not changed. Explain and propose fixes.

Diagnosis: discontinuous conduction. The signature is exact — correct behaviour loaded, over-speed unloaded, no change in the control signal.

The mechanism. A single-phase full converter provides only two current pulses per mains cycle, so there is a 10 ms gap between them. Under load, the armature inductance carries current across the gap and \( V_a = (2V_m/\pi)\cos\alpha \) holds. Unloaded, the current decays to zero partway through, the thyristors turn off, and the armature terminals float at the back-EMF for the rest of the interval.

The average terminal voltage therefore rises, the machine speeds up until its EMF is high enough to balance the new average, and the relationship between \( \alpha \) and speed becomes load-dependent — which is why it also becomes erratic.

Four remedies, in order of cost:

Add a series smoothing reactor. Doubling the circuit inductance roughly halves the current at which conduction becomes discontinuous. The traditional fix, and often the cheapest.
Change to a three-phase converter. Six pulses per cycle means 3.3 ms gaps instead of 10 ms, so continuous conduction extends to about a third of the load current. Usually decisive.
Provide a minimum load — a bleed resistor or a small mechanical drag. Crude, wasteful, but immediate.
Close the speed loop with a tachometer, if it is currently open-loop. The loop corrects the error regardless of its cause, though the gain change across the boundary must be allowed for.

The general point. This is Chapter 11's discontinuous conduction with a mechanical symptom attached. The same effect that made a chopper's output voltage rise at light load makes a DC drive over-speed — and recognising the shared mechanism saves a great deal of diagnosis.

Section 25-7

Summary & Formula Sheet

Chapter 25 in five sentences:

The commutator holds the armature and field fluxes perpendicular, so torque follows current and speed follows voltage independently — the property that made the DC drive dominant for a century.
Below base speed, vary the armature voltage for constant available torque; above it, weaken the field for constant power.
Field weakening is limited by commutation before torque, and the field's own time constant makes it slow to change.
A phase-controlled drive suits mains and regenerates naturally but loses regulation in discontinuous conduction; a chopper responds far faster and needs no smoothing reactor.
The current loop sits inside the speed loop so that clamping one signal gives torque limiting by construction.

Formula sheet · Chapter 25

Back-EMFthe machine's speed sensor

\( E_a = k\Phi\,\omega \)

Torquecurrent is torque, directly

\( T = k\Phi\,I_a \)

Armature circuitsteady state

\( V_a = E_a + I_aR_a \)

Speed–torque characteristica straight line

\( \omega = \dfrac{V_a}{k\Phi} - \dfrac{R_a}{(k\Phi)^2}T \)

Base speedboundary of the two regions

\( \omega_{base} = \dfrac{V_{rated} - I_{rated}R_a}{k\Phi_{rated}} \)

Field-weakening speedconstant power above base

\( \omega = \omega_{base}\dfrac{\Phi_{rated}}{\Phi} \)

Phase-controlled armature voltagethree-phase full converter

\( V_a = 1.35\,V_L\cos\alpha \)

Chopper-fed armature voltagecontinuous conduction

\( V_a = D\,V_s \)

Electrical time constantsets the current loop

\( \tau_e = L_a/R_a \)

Mechanical time constantsets the speed loop

\( \tau_m = \dfrac{JR_a}{(k\Phi)^2} \)

Plugging currentwhy a resistor is mandatory

\( I = \dfrac{V + E}{R_a + R_{plug}} \)

Braking energysizes the resistor's thermal mass

\( W = \tfrac12 J\omega^2 \)

Key terms

Separately-excited machine: Field and armature supplied independently, so flux and torque can be controlled separately. The standard drive configuration.
Base speed: Speed at rated voltage, rated flux and rated current. Below it, vary voltage; above it, weaken field.
Constant-torque region: Below base speed. Maximum torque is fixed; power rises with speed.
Constant-power region: Above base speed. Torque falls as the flux does; power stays at its rated value.
Field weakening: Reducing the field flux to raise speed above base. Limited by commutation, and slow because of the field time constant.
Flashover: An arc across the commutator, caused by excessive speed, current or field distortion. Destroys the machine quickly.
Cascade control: A fast inner current loop inside a slower outer speed loop. Gives torque limiting by clamping one signal.
Anti-windup: Preventing the speed controller's integrator accumulating while its output is saturated at the current limit.
Regenerative braking: Returning kinetic energy to the supply. Requires a converter that can invert and a supply that will accept it.
Dynamic braking: Dissipating kinetic energy in a resistor. Always works; wastes the energy; decelerates exponentially.
Plugging: Reversing the armature voltage while running. Fastest and most stressful; needs a current-limiting resistor and a zero-speed cut-out.

Check yourself

Test Yourself

Chapter 25 · six questions answers hidden until you ask

Why does weakening the field make a DC machine run faster, when weaker flux means weaker torque?

Because speed is set by the back-EMF balance, not by torque — and weaker flux means the machine must spin faster to generate the same EMF.

Follow the sequence. The armature circuit demands \( E_a = V_a - I_aR_a \), and \( V_a \) is fixed at its maximum. So \( E_a \) is essentially fixed too. But \( E_a = k\Phi\omega \), so:

\[ \omega = \frac{E_a}{k\Phi} \]

Halve \( \Phi \) and \( \omega \) must double to keep \( E_a \) where the circuit requires it.

What happens transiently, which makes it intuitive. Reduce the field and, for an instant, \( E_a \) drops. The armature circuit then sees a much larger net voltage \( V_a - E_a \), so the current surges, so the torque surges, so the machine accelerates — until the rising speed restores \( E_a \) and the current settles back.

The steady-state torque is unchanged if the load is unchanged — the machine simply draws more current to produce the same torque from less flux:

\[ T = k\Phi I_a \;\Longrightarrow\; \text{half } \Phi \text{ needs double } I_a \text{ for the same } T \]

Which is exactly why the available torque falls. The current is limited to \( I_{rated} \), so \( T_{max} = k\Phi I_{rated} \) halves. The machine can reach twice the speed but can only produce half the torque there — constant power.

And the practical caution. Because \( I_a \) must rise to hold torque, a field-weakened machine running a constant-torque load will overload. Field weakening suits loads whose torque falls with speed — winders, spindles, traction at cruise — and is unsuitable for constant-torque loads such as conveyors.

Why is the current loop inside the speed loop rather than the other way round?

Four reasons, and the second is the one that decides it.

1. Current is torque. \( T = k\Phi I_a \), so the speed loop's output is a torque demand expressed in amps — a physically meaningful, directly controllable quantity.

2. Torque limiting becomes a clamp rather than a trip. Limit the speed controller's output to \( I_{max} \) and the machine can never exceed it, whatever the speed error. During a hard start or a stall, the drive delivers maximum torque and holds it. Without this structure, protection would have to interrupt operation rather than limit it — and a stalled machine would draw \( V/R_a \), which for a 440 V, 0.4 Ω armature is 1100 A against a rated 100 A.

3. The time constants separate cleanly. Electrical: \( L_a/R_a \), a few milliseconds. Mechanical: \( JR_a/(k\Phi)^2 \), often hundreds. A fast inner loop and a slow outer loop are nearly independent, so each can be tuned alone — which is what makes commissioning tractable.

4. The inner loop hides the plant. Once the current loop is fast and closed, the speed loop sees a torque source. Supply variation, back-EMF and armature resistance are all rejected before the outer loop notices, so the speed loop's design does not depend on them.

Why the inverse makes no sense. A current loop outside a speed loop would ask the machine to hold a current by adjusting its speed — but the speed is set by the load, not by the drive's wishes. There is no actuator for it.

And the necessary companion: anti-windup. While the speed controller is saturated at the current limit, its integrator keeps accumulating error. Without clamping, the accumulated term causes a large overshoot when the speed finally arrives. Every practical drive freezes or clamps the integrator during saturation, and an overshoot on every start is the classic symptom of its absence.

A phase-controlled drive over-speeds when the load is removed, with no change in firing angle. Why?

Discontinuous conduction. The armature current reaches zero between firing pulses, so the terminals float at the back-EMF and the average voltage rises above \( 1.35V_L\cos\alpha \).

The mechanism. Between pulses the converter's instantaneous output can fall below the machine's back-EMF. Under load, enough current is flowing that the armature inductance carries it across the gap. Unloaded, the current decays to zero, the thyristors turn off, and the armature is disconnected from the converter for the remainder of the interval.

What that does to the average. During the disconnected interval the terminals sit at \( E_a \) rather than at the converter's output — and \( E_a \) is higher than the converter's average would be at that instant. So the mean terminal voltage rises, the machine accelerates, and it settles at whatever speed makes the EMF consistent with the new average.

Why it also becomes erratic. In continuous conduction the speed depends only on \( \alpha \). In discontinuous conduction it depends on \( \alpha \) and the load current, so any small load variation moves the speed — and the control loop's gain changes with it.

Remedies, in order of cost:

A series smoothing reactor. More inductance carries the current further between pulses. Traditional, and often cheapest.
Three-phase rather than single-phase. Six pulses per cycle means 3.3 ms gaps rather than 10 ms, extending continuous conduction to about a third of the current. Usually decisive on its own.
A minimum load — crude, wasteful, immediate.
Close the speed loop, which corrects the error regardless of cause, though the gain change across the boundary must be allowed for.

And note the shared mechanism. This is exactly Chapter 11's discontinuous conduction — where a chopper's output rose above \( DV_s \) at light load. Same physics, mechanical symptom. Recognising the family saves a lot of diagnosis.

Reversing the armature supply of a running machine draws 21 times rated current. Where does that come from?

From the back-EMF and the reversed supply adding, across nothing but the armature resistance.

The arithmetic. A machine at rated speed has \( E_a = 400 \) V, and the armature current is normally set by the small difference \( V_a - E_a = 40 \) V across \( R_a = 0.4\ \Omega \). Reverse \( V_a \) to −440 V and the two now oppose the current path in the same direction:

\[ I = \frac{V_a + E_a}{R_a} = \frac{440 + 400}{0.4} = 2100\ \text{A} \]

Twenty-one times rated. Three things fail almost immediately:

The commutator flashes over — an arc bridges the segments and, once established, the whole armature current flows through it.
The armature conductors are thrown out of their slots by \( F = BIl \) forces proportional to current.
The converter fails, since no reasonable device is rated for 21× its normal current.

So plugging always needs a series resistor, sized for about twice rated current:

\[ R_{total} = \frac{V + E}{2I_{rated}} = \frac{840}{200} = 4.2\ \Omega \;\Longrightarrow\; R_{plug} = 3.8\ \Omega \]

And the energy is worse than dynamic braking. During plugging the supply is also pushing power into the machine, so the resistor dissipates the kinetic energy plus the supplied energy — roughly twice what a dynamic brake would.

One more trap. The machine does not stop at zero speed — the reversed voltage accelerates it backwards. A zero-speed detector and a contactor are mandatory, and forgetting them turns an emergency stop into an emergency reversal.

Which is why plugging is for emergency stops only. For routine deceleration it is worse than either alternative in every respect except speed.

A drive is specified as "four-quadrant, fully regenerative", yet it trips on overvoltage when decelerating. Is the specification wrong?

No — the specification describes the converter, and regeneration also requires the system to accept the power. Those are separate questions.

What the converter can do. A dual converter or four-quadrant chopper can reverse the direction of power flow: the machine generates, and the converter passes that power back towards its supply. That capability is real.

What the system must do. The returned power has to be absorbed by something. If nothing will take it, it charges the supply-side capacitance and the voltage rises until protection operates.

Three situations where nothing absorbs it:

A diode rectifier front end. Diodes cannot conduct backwards, so the DC link simply charges. This is by far the most common case, and it is easy to miss because the inverter stage genuinely is four-quadrant.
A lightly-loaded site. The regenerating drive becomes the largest source present, and the local voltage rises.
A generator supply. An engine-driven alternator cannot accept reverse power at all — it would have to motor the engine.

The remedies:

A braking chopper across the DC link — a resistor switched in above a voltage threshold. Simple, reliable, wastes the energy. The usual answer.
An active front end — a PWM rectifier that returns power to the mains. Efficient and gives unity power factor, but roughly doubles the converter cost.
A shared DC link between several drives, so a decelerating machine feeds an accelerating one. Very effective where motions are complementary, and cheap.
A slower deceleration ramp, keeping the regenerated power within what the site absorbs. Free, and often sufficient.

The lesson generalises well beyond drives: a converter's capability and a system's behaviour are different claims, and a data sheet only ever makes the first.

The DC drive was displaced by AC drives from the 1990s. Where is it still specified?

In a small number of places where its remaining advantages outweigh the brushes.

What displaced it. Field-oriented control (Chapter 26) gave the induction motor the same decoupled torque and flux control, on a machine with no brushes, no commutator, a sealed enclosure and roughly half the cost per kilowatt. Once affordable DSPs made that practical, the argument was largely over.

Where it survives:

Existing installations. A working 500 kW rolling mill drive with decades of spares and a maintenance team who know it is not replaced for elegance. Much of the installed base will run out its life.
Very low-speed, very high-torque duty. A DC machine produces full torque at zero speed with no special measures. An induction motor can too, under FOC — but the DC solution is simpler and needs no encoder.
Small, cost-sensitive products. Below a few hundred watts, a permanent-magnet DC motor and a single transistor beat any AC alternative on cost. Cordless tools, small pumps and toys are all DC.
Applications wanting an inherently well-behaved characteristic. A series DC machine's torque rises steeply as speed falls, which suits traction and starting so naturally that it was the universal choice for railway and tram drives for eighty years.
Teaching and analysis. The DC machine remains the reference model — vector control is explicitly described as "making an induction motor behave like a DC machine", so the DC machine has to be understood first.

Where it is genuinely obsolete: general-purpose industrial drives above a few kilowatts, anything in a dusty or explosive atmosphere, anything requiring high speed, and anything where maintenance access is expensive.

The honest summary. The DC drive lost on maintenance, not on control. Everything it did well is now done as well or better by machines with no brushes — but the concepts it established are still how every modern drive is described.

Practice

Problems

Three habits for DC drive problems:

Find \( k\Phi \) first, from rated conditions. Everything else follows from it.
Check which region you are in. Below base speed the flux is rated; above it the voltage is.
For any braking question, ask where the energy goes before choosing a method.

Problems 1–5 are direct application; 6–9 need judgement; 10–12 are design questions worth discussing in a tutorial.

A 220 V, 50 A, 1200 rpm separately-excited machine has \( R_a = 0.3\ \Omega \). Find \( k\Phi \), the rated torque and the no-load speed at rated voltage.
For the machine of Problem 1, find the armature voltage needed for 600 rpm at rated torque, and the speed at rated voltage with half the rated torque.
The machine of Problem 1 must reach 2000 rpm. Find the flux ratio required, the torque available at rated current, and the shaft power.
A 440 V, 1500 rpm machine is fed from a three-phase full converter on a 415 V supply. Find the firing angle at rated speed and at 500 rpm with rated torque, and the displacement factor at each.
A chopper-fed drive runs from a 600 V DC traction line. Find the duty ratio for 400 V armature voltage, and the duty ratio at which the machine reaches base speed if \( I_a = 80 \) A and \( R_a = 0.25\ \Omega \).
Explain why the speed droop \( R_aT/(k\Phi)^2 \) worsens in the field-weakening region, and compute the droop at full flux and at 40% flux for a machine with \( R_a = 0.5\ \Omega \), \( k\Phi_{rated} = 3.0 \), at rated torque of 300 N·m.
A machine running at 1200 rpm with \( E = 200 \) V is to be plugged. With \( R_a = 0.25\ \Omega \) and a 220 V supply, find the resistance needed to limit the current to twice the rated 60 A, and the initial braking torque.
A 30 kW drive must stop 12 kg·m² from 1000 rpm in 6 s. Compare dynamic and regenerative braking on hardware, energy, deceleration profile and cost, and recommend one for (a) a machine stopping twice a day, (b) one stopping every 30 seconds.
A drive's speed overshoots by 25% on every start but tracks accurately once running. Diagnose the fault and state the fix.
Design the converter and control for a 75 kW, 500 V, 1000 rpm reversing mill drive on a 415 V supply: choose a converter topology, find the firing-angle range, specify the control structure, and state how braking energy will be handled.
A traction drive uses a series DC machine. Derive its speed–torque characteristic, explain why it suits traction, and state the one operating condition that must be prevented.
Compare a phase-controlled DC drive and a modern induction-motor drive for a 100 kW extruder running continuously at 40% of base speed. Consider efficiency, input power factor, harmonics, maintenance and capital cost, and recommend one with reasons.